SOLUTION: Please help me solve this word problem: A ball is thrown into the air and follows a path modeled by the function h(t)= -5(t-1.4)^2 +10, where time, t, is in seconds and height h(t

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Please help me solve this word problem: A ball is thrown into the air and follows a path modeled by the function h(t)= -5(t-1.4)^2 +10, where time, t, is in seconds and height h(t      Log On


   

Question 882158: Please help me solve this word problem: A ball is thrown into the air and follows a path modeled by the function h(t)= -5(t-1.4)^2 +10, where time, t, is in seconds and height h(t) is in meters. Determine when the ball will hit the ground.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
h%28t%29=+-5%28t-1.4%29%5E2+%2B10 is the height of the ball above the ground t seconds after it is thrown.
h%28t%29=0 means the ball is 0 meters above the ground, meaning that it is on the ground.
The equation making h%28t%29=0 gives you 2 answers for t , one negative and one positive.
The negative answer makes no sense, because we only care about the time after the ball was thrown.
(We do not know or care where the ball was for the "negative" values of t that represent before the throw. Maybe it was in someone's pocket before being thrown, but we do not care).

system%28h%28t%29=+-5%28t-1.4%29%5E2+%2B10%2Ch%28t%29=0%29 ---> -5%28t-1.4%29%5E2+%2B10=0 ---> 10=5%28t-1.4%29%5E2 ---> 10%2F5=%28t-1.4%29%5E2 ---> 2=%28t-1.4%29%5E2
From %28t-1.4%29%5E2=2 we get the two answers:
%28t-1.4%29%5E2=2--->system%28t-1.4=sqrt%282%29%2C%22or%22%2Ct-1.4=sqrt%282%29%29--->system%28t=sqrt%282%29%2B1.4%2C%22or%22%2Ct=-sqrt%282%29%2B1.4%29
highlight%28t=sqrt%282%29%2B1.4=about2.8%29 is the answer you want.
The ball hits the ground highlight%28about2.8%29 seconds after being thrown.